In Part-1 point 2 --- 5 mL of 0.9 M acetic acid is used
In Part-2 point 1 --- 15 mL of vinegar (nothing but acetic acid) is used
1 mol acetic acid reacts with 1 mol NaOH
from Part-1 we found volume of NaOH (sudden change in pH) = 3 mL for 5 mL of acid
hence for 15 mL of acid, NaOH will be 3 x 3 = 9 mL
total volume = 15 mL + 9 mL = 24 mL = 0.024 L
moles vinegar in 15 mL = 15 mL x 10-3 L x 0.9 M = 0.0135 moles
Molarity = moles / liter = 0.0135 moles / 0.024 L = 0.5625 M
Answers:
Molarity of vinegar = 0.5625 M
Volume of NaOH added = 9 mL
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